from __future__ import division
from sympy import *
init_printing(use_unicode=True)

import numpy as np

# matplotlib
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
from matplotlib.ticker import LinearLocator, FormatStrFormatter
%matplotlib inline
plt.style.use('seaborn-white')

## plotly
from plotly.offline import init_notebook_mode, iplot 
import plotly.graph_objs as go
import plotly.offline as py
from plotly.subplots import make_subplots
import plotly.express as px
# Graphics in retina format 

## Text
from IPython.display import display, Markdown, Latex

import warnings
warnings.filterwarnings("ignore")

Home / Linear Algebra

Python Lab

• Getting Started with Python
• Chapter 1: Systems of Equations
• Chapter 2: Matrices
• Chapter 3: Determinants
• Chapter 4: Rn
• Chapter 5: Linear Transformations
• Chapter 6: Complex Numbers
• Chapter 7: Spectral Theory
• Chapter 7: Markov Matrices

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Python Lab: Systems of Equations

Linear equation: If $a$, $b$, and $c$ are real numbers, the graph of an equation of the form $$ax+by = c$$ is a straight line (if $a$ and $b$ are not both zero), so such an equation is called a linear equation in the variables $x$ and $y$.

Example: Solve \begin{cases} x+y=0\\ x-y=0,\\ \end{cases} Graphically.

matplotlib:

x = np.linspace( -10 , 10 ,100)
fig, ax = plt.subplots(nrows=1, ncols=1, figsize=(6, 6), sharex = False)
_ = ax.plot(x, -x, color='r', label='x + y = 0')
_ = ax.plot(x, x, color='b', label='x - y = 0')
_ = ax.set_xlabel('$x$', fontsize=18)
_ = ax.set_ylabel('$y$', fontsize=18)
_ = ax.legend(loc='best', bbox_to_anchor=(1.01, 0., 0.2, 0.6))
_ = ax.set_xlim([min(x),max(x)])
_ = ax.set_ylim([min(x),max(x)])
_ = ax.grid(True, which='both')
_ = ax.axhline(y=0, color='k')
_ = ax.axvline(x=0, color='k')

plotly:

x = np.linspace( -10 , 10 ,100)
fig = go.Figure()
fig.add_trace(go.Scatter(x= x, y= -x, line=dict(color='Blue', width= 1.5), name = 'x + y = 0'))
fig.add_trace(go.Scatter(x= x, y= x, line=dict(color='OrangeRed', width= 1.5), name = 'x - y = 0'))
            
# Update
fig.update_layout(plot_bgcolor= 'white', width = 600)
fig.update_xaxes(showline=True, linewidth=1, linecolor='Lightgray', mirror=True,
                 showgrid=True, gridwidth=1, gridcolor='Lightgray',
                 zeroline=True, zerolinewidth=1.5, zerolinecolor='black')
fig.update_yaxes(showline=True, linewidth=1, linecolor='Lightgray', mirror=True,
                 showgrid=True, gridwidth=1, gridcolor='Lightgray',
                 zeroline=True, zerolinewidth=1.5, zerolinecolor='black')
fig.update_xaxes(title_text='x', range=[-10, 10])
fig.update_yaxes(title_text='y', range=[-10, 10])
fig.show()

Example: Solve \begin{cases} x+y+z=2\\ x+y+z=5 \end{cases}.

matplotlib:

fig = plt.figure(figsize = (20, 10))
ax = fig.gca(projection='3d')

# Make data.
X = np.linspace( -10 , 10 ,100)
Y = np.linspace( -10 , 10 ,100)
X, Y = np.meshgrid(X, Y)
Z1 = 2 - X - Y 
Z2 = 5 - X - Y 

# Plot the surface.
surf = ax.plot_surface(X, Y, Z1, color = 'blue', linewidth=0, antialiased=False)
_ = ax.plot_surface(X, Y, Z2, color = 'lightgreen', linewidth=0, antialiased=False)
_ = ax.set_zlim(-20, 40)
_ = ax.zaxis.set_major_locator(LinearLocator(10))
_ = ax.set_xlabel('x', fontsize=18)
_ = ax.set_ylabel('y', fontsize=18)
_ = ax.set_zlabel('z', fontsize=18)

plotly:

X = np.linspace( -10 , 10 ,100)
Y = np.linspace( -10 , 10 ,100)
X, Y = np.meshgrid(X, Y)
Z1 = 2 - X - Y 
Z2 = 5 - X - Y 

fig = go.Figure()
fig.add_trace(go.Surface(z=Z1, x=X, y=Y, colorscale='Greens', colorbar_len=0.5))
fig.add_trace(go.Surface(z=Z2, x=X, y=Y, colorscale='Blues', colorbar_len=0.5))

fig.update_layout(scene = dict(xaxis = dict(tickvals= list(np.arange(-10,11,5))),
                               yaxis = dict(tickvals= list(np.arange(-10,11,5))),
                               zaxis = dict(nticks=5, ticks='outside',tick0=0, tickwidth=4),),
                  width=700, margin=dict(r=10, l=10, b=10, t=10))
fig.show()

Solving linear system using back-substitution

Example: Solve the following linear system using back-substitution. \begin{equation*} \begin{cases} x_{1}+2\,x_{2}+3\,x_{3}=2\\ 4\,x_{1}-3\,x_{2}+x_{3}=-3\\ x_{2}-2\,x_{1}+3\,x_{3}=5 \end{cases} \end{equation*}

The corresponding augmented matrix is

M = Matrix([[1 , 2 , 3 , 2], [4 , -3 , 1 , -3], [-2 , 1 , 3 , 5]]) 
M

$\displaystyle \left[\begin{matrix}1 & 2 & 3 & 2\\4 & -3 & 1 & -3\\-2 & 1 & 3 & 5\end{matrix}\right]$

In RREF:

RREF= M.rref()[0]
RREF

$\displaystyle \left[\begin{matrix}1 & 0 & 0 & -1\\0 & 1 & 0 & 0\\0 & 0 & 1 & 1\end{matrix}\right]$

Therefore, the solutions are:

display(Latex(r'$x_1$ = %i, $x_2$ = %i and $x_3$ = %i' % (RREF[0,-1], RREF[1,-1], RREF[2,-1])))

$x_1$ = -1, $x_2$ = 0 and $x_3$ = 1

To verify our answer, we can use linalg from Numpy. This requires converting the Matrix M into an array.

A = np.asarray(M[:,:-1], dtype = int)
B = np.asarray(M[:,-1], dtype = int)
Matrix(np.linalg.solve(A, B))

$\displaystyle \left[\begin{matrix}-1.0\\0.0\\1.0\end{matrix}\right]$

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